Question 272: Ray Optics

Question

A collimated beam of light of diameter 2 mm is propagating along $x$-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm , then the focal length of second lens is $\_\_\_\_$ mm.

Accepted Answer

For a beam to enter collimated (parallel rays) and exit collimated with a larger diameter, the two lenses must be placed such that their focal points coincide.
The first lens focuses the parallel rays at its focal point ( $f_1$ ) and then the second lens is placed so that its focal point $\left(f_2ight)$ is at the same location. It then re-collimates the rays diverging from that point.
From similar triangles in the ray diagram, the ratio of the diameters of the beams is directly proportional to the ratio of the focal lengths of the lenses.
$$ \frac{D_2}{D_1}=\frac{f_2}{f_1} $$
Where :
$\mathrm{D}_1=$ Initial beam diameter $=2 \mathrm{~mm}$
$\mathrm{D}_2=$ Final beam diameter $=14 \mathrm{~mm}$
$\mathrm{f}_1=$ Focal length of the first lens $=40 \mathrm{~mm}$
$f_2=$ Focal length of the second lens (need to be calculated)
$$\Rightarrow $$$\mathrm{f}_2=\mathrm{f}_1 	imes\left(\frac{\mathrm{D}_2}{\mathrm{D}_1}ight)$
$$\Rightarrow $$ $\mathrm{f}_2=40 	imes\left(\frac{14}{2}ight)$
$$\Rightarrow $$ $$ \mathrm{f}_2=280 \mathrm{~mm} $$
Therefore, to expand the beam from a diameter of 2 mm to 14 mm using a first lens of 40 mm focal length, the second lens must have a focal length of 280 mm .
Hence, the correct answer is 280 .

Ray Optics

A collimated beam of light of diameter 2 mm is propagating along x-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm , then the focal length of second lens is \_\_\_\_ mm.

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